How do you find the vertex and the intercepts for #f(x) = -x^2 + x - 5#?

1 Answer
Andy Y.
May 17, 2017

Vertex: #=(1/2, 5 3/4)#
#y#-intercept: #x=0#, #y=-5#
#x#-Intercept: none

Explanation:

This equation is in standard form
#y=ax^2+bx+c

Where #a=-1#, #b=1#, and #c=-5#

Vertex: #(-b/2a, f (-b/2a))#
#=(1/2, 5 3/4)#

Intercepts:
When #x=0#, #y=-5#
When #y=0#, you have to use the quadratic equation. But the number under the square root is negative, so there are no #x#-Intercepts.

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